Summer 2015 Math Practice Problems SOLUTIONS SCHOOL OF PUBLIC POLICY AND GOVERNANCE Math practice...

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EVANS SCHOOL OF PUBLIC POLICY AND GOVERNANCE Math practice problems for incoming MPA students Summer 2015 SOLUTIONS 1. INVERTING EQUATIONS a) = + 4 = βˆ’ b) = 2 + 4 = βˆ’ c) = 3 + 5 = βˆ’ d) = 4 βˆ’ 8 = + e) = .25 βˆ’ 100 = + f) = βˆ’ ! ! βˆ’ 100 = βˆ’ βˆ’ g) = βˆ’.1 + 400 = βˆ’ + h) = 32 + ! ! = βˆ’ 2. SOLVING PAIRS OF SIMULTANEOUS EQUATIONS a) 3 + = 13 + 6 = βˆ’ 7 = βˆ’, = b) = + 4 = 2 + 4 = βˆ’, = βˆ’ c) 5 = 2 + 3 2 βˆ’ = 0 = , = d) + 2 = 8 βˆ’ 2 = 4 = , = e) 4 + = 9 βˆ’ = 1 = , = f) 2 + 3 = 28 + = 11 = , = g) 11 + 6 = 79 11 + 3 = 67 = , = h) ! ! + ! ! = 8 2 3 + 3 2 = 17 = , = 3. GRAPHING FROM AN EQUATION OR ITS INVERSE a) Graph = 3 b) Graph = + 3 c) Graph = ! ! + 1

Transcript of Summer 2015 Math Practice Problems SOLUTIONS SCHOOL OF PUBLIC POLICY AND GOVERNANCE Math practice...

EVANS SCHOOL OF PUBLIC POLICY AND GOVERNANCE Math practice problems for incoming MPA students Summer 2015

SOLUTIONS

1. INVERTING EQUATIONS a) 𝑦 = π‘₯ + 4 Γ  𝒙 = π’š βˆ’ πŸ’

b) π‘₯ = 2𝑦 + 4 Γ   π’š = 𝟏

πŸπ’™ βˆ’ 𝟐

c) 𝑦 = 3π‘₯ + 5 Γ  𝒙 = 𝟏

πŸ‘π’š βˆ’ πŸ“

πŸ‘

d) 𝑦 = 4π‘₯ βˆ’  8 Γ  𝒙 = 𝟏

πŸ’π’š + 𝟐

e) 𝑃 = .25𝑄 βˆ’ 100 Γ  𝑸 = πŸ’π‘· + πŸ’πŸŽπŸŽ

f) 𝑦 = βˆ’ !!π‘₯ βˆ’  100 Γ  𝒙 = βˆ’πŸ’π’š βˆ’  πŸ’πŸŽπŸŽ

g) 𝑃 = βˆ’.1𝑄 +  400 Γ  𝑸 = βˆ’πŸπŸŽπ‘· +  πŸ’πŸŽπŸŽπŸŽ

h) 𝑦 = 32 + !

!π‘₯ Γ  𝒙 = πŸ“

πŸ–π’š βˆ’ 𝟐𝟎

2. SOLVING PAIRS OF SIMULTANEOUS EQUATIONS a) 3π‘₯ + 𝑦 = 13

π‘₯ + 6𝑦 = βˆ’  7 π’š =  βˆ’πŸ, 𝒙 = πŸ“

b) 𝑦 = π‘₯ + 4 π‘₯ = 2𝑦 + 4 π’š =  βˆ’πŸ–, 𝒙 = βˆ’πŸπŸ

c) 5π‘Ž = 2𝑏 + 3 2π‘Ž βˆ’ 𝑏 = 0 𝒂 =  πŸ‘,𝒃 = πŸ”

d) π‘₯ + 2𝑦 = 8 π‘₯ βˆ’ 2𝑦 = 4 π’š =  πŸ, 𝒙 = πŸ”

e) 4π‘₯ + 𝑦 = 9 π‘₯ βˆ’ 𝑦 = 1 π’š =  πŸ, 𝒙 = 𝟐

f) 2π‘₯ + 3𝑦 =  28 π‘₯ + 𝑦 =  11 π’š =  πŸ”, 𝒙 = πŸ“

g) 11π‘₯   +  6𝑦 = 79

11π‘₯ + 3𝑦 = 67 π’š =  πŸ’, 𝒙 = πŸ“

h) !!π‘₯   +  !

!𝑦 = 8

23π‘₯   +  

32𝑦 = 17

π’š =  πŸ”, 𝒙 = 𝟏𝟐

3. GRAPHING FROM AN EQUATION OR ITS INVERSE a) Graph 𝑦 = 3π‘₯ b) Graph 𝑦 = π‘₯ + 3 c) Graph 𝑦 = !

!π‘₯ + 1

d) Graph 𝑦 = !!π‘₯ + 8 e) Graph π‘₯ = 𝑦 + 3 f) Graph π‘₯ = !

!𝑦 + 1

g) Graph 𝑦 = 5π‘₯ and 𝑦 = βˆ’5π‘₯ + 50 h) Graph π‘₯ = !

!𝑦 βˆ’ 7.5 and 𝑦 = βˆ’ !

!π‘₯ + 25

4. RECOVERING THE EQUATION OF A LINE FROM A GRAPH a) Find the equation for the line. b) Find the equation for the line. c) Find the equation for the line.

π’š = βˆ’ 𝟐

πŸ‘π’™ + 𝟐𝟏 π’š = βˆ’ 𝟏

πŸπŸŽπ’™ + πŸ— π’š = πŸ‘π’™ + 𝟏𝟎

d) Find the equation for the line. e) Find the equation for the line. f) Find the equation for the line.

π’š = βˆ’ πŸπŸ“

πŸ—π’™ + πŸπŸ“ π’š = πŸ“

πŸπŸ•π’™ π’š = 𝟏

πŸ—π’™

a) Find the equation for the line. b) Find the equation for the line. c) Find the equation for the line.

π’š = βˆ’ 𝟐

πŸ—π’™ + πŸπŸ• π’š = πŸ“ 𝒙 = πŸπŸ“

5. CALCULATING AREAS UNDER STRAIGHT LINE CURVES

a) Find the area under the line. b) Find the area under the line. c) Find the area under the line.

(21 X 4) / 2 = 42 (35 X 2) / 2 = 35 (15 X 27) / 2 = 202.5

d) Find the area under the line. e) Find the area under the line. f) Find the area under the line.

((21-6) X 6) / 2 + (6 X 6) = 81 ((40-15) X 9) / 2 + (15 X 9) = 247.5 ((27-6) X 6) / 2 + (6X6) = 99

g) Find the area under the line. h) Find the area under the line. i) Find the area under the line.

(12 X 4) / 2 + ((21-12) X (10-4)) / 2 (30-5)X6 / 2 + (9-6)X5) / 2 ((12-6)X12)/2 + ((15-12) X (24-12)) + (12 X (10-4)) = 123 + (5 X 6) = 112.5 / 2 + (6X12) + (24-12) X 12 = 270

6. NATURAL LOGS AND EXPONENTIAL FUNCTIONS

a) Express ln 2.7183 = 1 in exponential form. Γ  Solution: 2.7183 = e1 b) Write the equation e2.7 β‰ˆ 14.88 in logarithmic form. Γ  Solution: ln 14.88 = 2.7 c) Express the equation e2 β‰ˆ 7.39 in logarithmic form. Γ  Solution: ln 7.39 = 2 d) Write as a single logarithm: ln 3 + ln 7 Γ  Solution: ln 21 (using the product property) e) Write as a single logarithm: ln 6 – ln 2 Γ  Solution: ln 3 (using the quotient property) f) Expand the following expression: ln 12π‘₯! Γ  Solution: π₯𝐧 𝟏𝟐 + πŸ’ π₯𝐧 𝒙 (using the product and power

properties) g) Expand the following expression: ln !!

!

!! Γ  Solution: ln 4 + 3 ln y – 5 ln x

h) Solve for x: ln x = 24 Γ  Solution: x = e24 i) Solve for x: e4x+2 = 50 Γ  Solution: x = .478 j) Solve 1 + 2e1-3z = 15. Γ  z = -.315